Chapter 10.2, Problem 44E

To determine

To sketch: The path that the particle travels.

Answer to Problem 44E

The answer:

The path of the particle between $t=0$ and

  $t=3$ is the parametric curve defined by

  $x(t)=\sin (4\pi t)/2+7$

and

  $y(t)=-2\cos (2\pi t)+4$

For

  $0\le t\le 3$

The following graph is the graph of the path of the particle between $t=0$ and

  $t=3$ is shown below:

  

Explanation of Solution

Given information:

The velocity of $v(t)=3t^2-2t,1+\cos \pi t;(2,6)$ .

Calculation:

Know that if the position vector of a particle at time $t$ is equal to:

  $r(t)=(x(t),y(t))$

The velocity vector of a particle at time $t$ is equal to:

  $v(t)=\frac{dx}{dt},\frac{dy}{dt}$

From exercise $40$ , the velocity vector of a particle at time $t$ is equal to:

  $v(t)=(2\pi \cos (4\pi t),4\pi \sin (2\pi t))$

Which means that,

  $\begin{array}{c}dx/dt=2\pi \cos (4\pi t)\\ dy/dt=4\pi \sin (2\pi t)\end{array}$

Then,

  $\begin{array}{c}x(t)={{\displaystyle \int }}^{\text{}}\frac{dx}{dt}dt\\ ={{\displaystyle \int }}^{\text{}}2\pi \cos (4\pi t)dt\\ =2\pi \cdot \frac{\sin (4\pi t)}{4\pi }+C_x\\ =\frac{\sin (4\pi t)}{2}+C_x\\ y(t)={{\displaystyle \int }}^{\text{}}\frac{dy}{dt}dt\\ ={{\displaystyle \int }}^{\text{}}4\pi \sin (2\pi t)dt\\ =4\pi \cdot \frac{-\cos (2\pi t)}{2\pi }+C_y\\ =-2\cos (2\pi t)+C_y\end{array}$

To calculate the integral:

  $\begin{array}{c}{{\displaystyle \int }}^{\text{}}\cos (4\pi t)dt=\frac{1}{4\pi }{{\displaystyle \int }}^{\text{}}\cos udu\\ =\frac{\sin u}{4\pi }+C\\ =\frac{\sin (4\pi t)}{4\pi }+C\end{array}$

To calculate the integral:

  $\begin{array}{c}{{\displaystyle \int }}^{\text{}}\sin (2\pi t)dt=\frac{1}{2\pi }{{\displaystyle \int }}^{\text{}}\sin udu\\ =-\frac{\cos u}{2\pi }+C\\ =-\frac{\cos (2\pi t)}{2\pi }+C\end{array}$

From exercise $40$ , the position of the particle at time $t=0$ is equal to $(7,2)$ , which means that,

  $\begin{array}{c}x(0)=7\\ y(0)=2\end{array}$

Then,

  $\begin{array}{c}x(0)=\frac{\sin (4\pi \cdot 0)}{2}+C_x\\ =\frac{\sin (0)}{2}+C_x\\ =\frac{0}{2}+C_x=0+C_x\\ =C_x\\ \Rightarrow C_x=7\\ y(0)=-2\cos (2\pi \cdot 0)+C_y\\ =-2\cos (0)+C_y\\ =-2\cdot 1+C_y\\ =C_y-2\\ \Rightarrow C_y=4\end{array}$

According to the previous result,

  $\begin{array}{c}x(t)=\sin (4\pi t)/2+7\\ y(t)=-2\cos (2\pi t)+4\end{array}$

Which means that,

  $r(t)=\frac{\sin (4\pi t)}{2}+7,-2\cos (2\pi t)+4$

The path of the particle between $t=0$ and

  $t=3$ is the parametric curve defined by

  $x(t)=\sin (4\pi t)/2+7$

and

  $y(t)=-2\cos (2\pi t)+4$

For

  $0\le t\le 3$

The following graph is the graph of the path of the particle between $t=0$ and

  $t=3$ is shown below: